Q.

The perpendicular distance, of the line x–12=y+2–1=z+32 from the point P(2, –10, 1) is:          [2025]

1 6  
2 35  
3 52  
4 43  

Ans.

(2)

x–12=y+2–1=z+32=λ (say)

Any point on the line is given by

M(x,y,z)≡(2λ+1,–λ–2,2λ–3)

Also, PM→=(2λ+1–2)i^+(–λ–2+10)j^+(2λ–3–1)k^

=(2λ–1)i^+(–λ+8)j^+(2λ–4)k^          ... (i)

∵  PM→·n→=0

⇒ (2λ–1)2+(–λ+8)(–1)+(2λ–4)2=0

⇒ 4λ–2+λ–8+4λ–8=0

⇒ 9λ–18=0 ⇒ λ=2

∴  |PM→|=32+62+02=45=35          [Using (i)]